3.342 \(\int \frac{x^m}{(a+b x^2)^2 (c+d x^2)^3} \, dx\)

Optimal. Leaf size=325 \[ \frac{d^2 x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-12 m+35\right )\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c^3 (m+1) (b c-a d)^4}+\frac{d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{8 a c^2 \left (c+d x^2\right ) (b c-a d)^3}-\frac{b^3 x^{m+1} (a d (7-m)-b (c-c m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^4}+\frac{d x^{m+1} (a d+2 b c)}{4 a c \left (c+d x^2\right )^2 (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

(d*(2*b*c + a*d)*x^(1 + m))/(4*a*c*(b*c - a*d)^2*(c + d*x^2)^2) + (b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(
c + d*x^2)^2) + (d*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m))*x^(1 + m))/(8*a*c^2*(b*c - a*d)^3*(c + d*x
^2)) - (b^3*(a*d*(7 - m) - b*(c - c*m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2
*a^2*(b*c - a*d)^4*(1 + m)) + (d^2*(b^2*c^2*(35 - 12*m + m^2) - 2*a*b*c*d*(7 - 8*m + m^2) + a^2*d^2*(3 - 4*m +
m^2))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^4*(1 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.613702, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {472, 579, 584, 364} \[ \frac{d^2 x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-12 m+35\right )\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c^3 (m+1) (b c-a d)^4}+\frac{d x^{m+1} \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )}{8 a c^2 \left (c+d x^2\right ) (b c-a d)^3}-\frac{b^3 x^{m+1} (a d (7-m)-b (c-c m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^4}+\frac{d x^{m+1} (a d+2 b c)}{4 a c \left (c+d x^2\right )^2 (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/((a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

(d*(2*b*c + a*d)*x^(1 + m))/(4*a*c*(b*c - a*d)^2*(c + d*x^2)^2) + (b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(
c + d*x^2)^2) + (d*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m))*x^(1 + m))/(8*a*c^2*(b*c - a*d)^3*(c + d*x
^2)) - (b^3*(a*d*(7 - m) - b*(c - c*m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2
*a^2*(b*c - a*d)^4*(1 + m)) + (d^2*(b^2*c^2*(35 - 12*m + m^2) - 2*a*b*c*d*(7 - 8*m + m^2) + a^2*d^2*(3 - 4*m +
m^2))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^4*(1 + m))

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 584

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx &=\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{x^m \left (2 a d-b c (1-m)-b d (5-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)}\\ &=\frac{d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac{\int \frac{x^m \left (2 \left (8 a b c d-2 b^2 c^2 (1-m)-a^2 d^2 (3-m)\right )-2 b d (2 b c+a d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2}\\ &=\frac{d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac{\int \frac{x^m \left (2 \left (24 a b^2 c^2 d-4 b^3 c^3 (1-m)-a^2 b c d^2 \left (11-12 m+m^2\right )+a^3 d^3 \left (3-4 m+m^2\right )\right )-2 b d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac{d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac{\int \left (\frac{8 b^3 c^2 (-b c (1-m)+a d (7-m)) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac{2 a d^2 \left (-b^2 c^2 \left (35-12 m+m^2\right )+2 a b c d \left (7-8 m+m^2\right )-a^2 d^2 \left (3-4 m+m^2\right )\right ) x^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac{d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac{\left (b^3 (b c (1-m)-a d (7-m))\right ) \int \frac{x^m}{a+b x^2} \, dx}{2 a (b c-a d)^4}+\frac{\left (d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right )\right ) \int \frac{x^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^4}\\ &=\frac{d (2 b c+a d) x^{1+m}}{4 a c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac{d \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right ) x^{1+m}}{8 a c^2 (b c-a d)^3 \left (c+d x^2\right )}-\frac{b^3 (a d (7-m)-b (c-c m)) x^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{2 a^2 (b c-a d)^4 (1+m)}+\frac{d^2 \left (b^2 c^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+a^2 d^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{d x^2}{c}\right )}{8 c^3 (b c-a d)^4 (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.159945, size = 54, normalized size = 0.17 \[ \frac{x^{m+1} F_1\left (\frac{m+1}{2};2,3;\frac{m+1}{2}+1;-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{a^2 c^3 (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m/((a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

(x^(1 + m)*AppellF1[(1 + m)/2, 2, 3, 1 + (1 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(a^2*c^3*(1 + m))

________________________________________________________________________________________

Maple [F]  time = 0.072, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( b{x}^{2}+a \right ) ^{2} \left ( d{x}^{2}+c \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x)

[Out]

int(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b^{2} d^{3} x^{10} +{\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{8} +{\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{6} + a^{2} c^{3} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} +{\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

integral(x^m/(b^2*d^3*x^10 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 + a^2*c
^3 + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(b*x**2+a)**2/(d*x**2+c)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")

[Out]

integrate(x^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)